101_Physics ProblemsTechnical Physics

101_Physics ProblemsTechnical Physics - 102 P4.50 Motion in...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
102 Motion in Two Dimensions P4.50 (a) yx g v x fi f ii f =− tan cos θ bg di 2 22 2 Setting xd f = cos φ , and yd f = sin , we have dd g v d i sin tan cos cos cos φθ b g b g 2 2 . FIG. P4.50 Solving for d yields, d v g iii i = 2 2 2 cos sin cos sin cos cos θθφ or d v g = 2 2 2 cos sin cos θθ . (b) Setting dd d i = 0 leads to i + 45 2 and d v g i max sin cos = 2 2 1 . P4.51 Refer to the sketch: (b) xv t xi = ; substitution yields 130 35 0 vt i cos . . yv t a t yi =+ 1 2 2 ; substitution yields 20 0 35 0 1 2 980 2 .s i n . . + t i bgaf . Solving the above gives t = 381 . s .
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online